Odpowiedź:
Szczegółowe wyjaśnienie:
a)
[tex]x^2 - 9 > 0\\(x - 3)(x + 3) > 0\\[/tex]
czyli
x∈ (-∞,-3)∪(3,+∞)
b)
[tex]6x^2 + 3x \geq 0\\3x(2x + 1)\geq 0\\\\x_1 = 0 \\2x + 1 = 0\\x = -\frac{1}{2}\\[/tex]
x∈(-∞,-1/2> ∪ <0,+∞)
c)
[tex]2x^2 + 5x -3 > 0\\\Delta = 25 - 4 * 2 * (-3) = 25 + 24 = 49\\\sqrt{\Delta} = 7\\x_1 = \frac{-5-7}{2*2} = \frac{-12}{4} = -3\\x_2 = \frac{-5+7}{2*2}= \frac{2}{4} = \frac{1}{2}[/tex]
x∈(-∞,-3)∪([tex]\frac{1}{2}[/tex], +∞)
